Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON3):

Solution:
n=int(input())
p=list(map(float,input().split()))
dp=[p[0],1-p[0]]
if n==1:
    print(p[0])
    exit()
for i in range(n-1):
    tmp=[]
    tmp.append(dp[0]*p[i+1])
    for j in range(i+1):
        tmp.append(dp[j]*(1-p[i+1])+dp[j+1]*p[i+1])    
    tmp.append(dp[-1]*(1-p[i+1]))
    dp=tmp
print(sum(dp[:n//2+1]))